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Exponentials and logarithms — solutions

21 full solutions · spec MM5.1-MM5.3

Section 1 Part 1: AS-level pure mathematics21 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. LOG-01Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Find the value of \(\log_2 40 - \log_2 5 + \log_2 \dfrac{1}{4}\).

  1. A\(1\)
  2. B\(2\)
  3. C\(-1\)
  4. D\(5\)
  5. E\(3\)
  6. F\(0\)
Answer: A. \(\log_2\left(\frac{40}{5}\cdot\frac14\right) = \log_2 2 = 1\).
Full method — how and why
Combine everything into a single logarithm using the laws, then evaluate.
  1. Combine the first two: \(\log_2 40 - \log_2 5 = \log_2\dfrac{40}{5} = \log_2 8\).
    A difference of logs (same base) is the log of the quotient.
  2. Add the third: \(\log_2 8 + \log_2\frac14 = \log_2\left(8 \times \frac14\right) = \log_2 2\).
    A sum of logs is the log of the product. Note \(\log_2\frac14\) is negative, so this step reduces the total.
  3. Evaluate: \(\log_2 2 = 1\).
    \(\log_a a = 1\) for any valid base.
Where it goes wrong
  • Treating \(\log_2\frac14\) as \(+2\) rather than \(-2\) and answering \(5\).
  • Dividing the logarithms instead of subtracting them.
Takeaway \(\log\) turns \(\times\) into \(+\) and \(\div\) into \(-\); combine into one log before evaluating.
2. LOG-02Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

Solve \(3^{2x} = 4\). Give \(x\) in terms of logarithms to base \(3\).

  1. A\(x = \log_3 4 - 2\)
  2. B\(x = \log_3 2\)
  3. C\(x = \dfrac{\log_3 4}{4}\)
  4. D\(x = \log_2 3\)
  5. E\(x = \dfrac{1}{2}\log_3 2\)
  6. F\(x = 2\log_3 2\)
Answer: B. \(2x = \log_3 4 = 2\log_3 2\), so \(x = \log_3 2\).
Full method — how and why
Solving \(a^{kx} = b\). Take logs to the given base and simplify the right-hand side.
  1. Take \(\log_3\) of both sides: \(\log_3 3^{2x} = \log_3 4\), so \(2x = \log_3 4\).
    \(\log_3 3^{u} = u\) - logs and exponentials with the same base undo each other.
  2. Rewrite the right-hand side: \(\log_3 4 = \log_3 2^2 = 2\log_3 2\).
    The power law \(\log a^k = k\log a\) brings the exponent down.
  3. Divide by \(2\): \(x = \log_3 2\).
    The factor \(2\) cancels exactly, which is why the answer is so clean.
Where it goes wrong
  • Stopping at \(x = \frac12\log_3 4\) and then mis-simplifying to \(\frac12 \log_3 2\).
  • Writing \(\log_3 4 = \log_3 2 \times \log_3 2\).
Takeaway \(\log a^k = k \log a\); rewriting numbers as powers often makes fractions cancel.
3. LOG-03Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

Solve \(2^{2x} - 3\cdot 2^{x+2} + 32 = 0\). Find the sum of the solutions.

  1. A\(6\)
  2. B\(12\)
  3. C\(4\)
  4. D\(3\)
  5. E\(8\)
  6. F\(5\)
Answer: F. Let \(y = 2^x\): \(y^2 - 12y + 32 = (y - 4)(y - 8) = 0\), so \(2^x = 4\) or \(8\): \(x = 2\) or \(3\). Sum \(5\).
Full method — how and why
A hidden quadratic in \(2^x\). The middle term needs rewriting before the substitution is visible.
  1. Rewrite the middle term: \(3 \cdot 2^{x+2} = 3 \cdot 2^2 \cdot 2^x = 12 \cdot 2^x\).
    \(2^{x+2} = 2^x \times 4\) - separating the constant part is essential before substituting.
  2. Substitute \(y = 2^x\), noting \(2^{2x} = y^2\): \(y^2 - 12y + 32 = 0\).
    \(2^{2x} = (2^x)^2\); that is what makes it quadratic.
  3. Factorise: \((y-4)(y-8) = 0\), so \(y = 4\) or \(y = 8\).
    Both are positive, so both give solutions for \(x\).
  4. Convert back: \(2^x = 4 \Rightarrow x = 2\); \(2^x = 8 \Rightarrow x = 3\). Sum \(= 5\).
    The question asks for the sum of the \(x\) values, not the \(y\) values.
Where it goes wrong
  • Summing the \(y\) values (\(4 + 8 = 12\)) - which is offered.
  • Reading \(2^{x+2}\) as \(2^x + 2\).
Takeaway \(a^{x+k} = a^k a^x\): split off the constant before substituting, then convert the roots back to \(x\).
4. LOG-04Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Given that \(\log_a x = 3\) and \(\log_a y = 5\), find \(\log_a \left(\dfrac{x^2 \sqrt{y}}{a}\right)\).

  1. A\(8\)
  2. B\(\dfrac{15}{2}\)
  3. C\(\dfrac{13}{2}\)
  4. D\(\dfrac{17}{2}\)
  5. E\(10\)
  6. F\(\dfrac{11}{2}\)
Answer: B. \(2\log_a x + \frac12\log_a y - \log_a a = 6 + \frac52 - 1 = \frac{15}{2}\).
Full method — how and why
Break the expression apart with the log laws, then substitute the given values.
  1. Split the quotient and product: \(\log_a\dfrac{x^2\sqrt y}{a} = \log_a x^2 + \log_a \sqrt y - \log_a a\).
    Products become sums, quotients become differences.
  2. Bring the powers down: \(2\log_a x + \frac12\log_a y - \log_a a\).
    \(\sqrt y = y^{1/2}\), so its log is \(\frac12\log_a y\).
  3. Substitute \(\log_a x = 3\), \(\log_a y = 5\), \(\log_a a = 1\): \(6 + \frac52 - 1\).
    \(\log_a a = 1\) always - a fact worth having instantly available.
  4. Add: \(\frac{12 + 5 - 2}{2} = \frac{15}{2}\).
    Put everything over \(2\).
Where it goes wrong
  • Forgetting \(\log_a a = 1\) and answering \(\frac{17}{2}\).
  • Using \(\log_a\sqrt y = \sqrt{\log_a y}\).
Takeaway Expand fully with the three laws before substituting; \(\log_a a = 1\) and \(\log_a 1 = 0\).
5. LOG-05Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_3(x + 6) + \log_3(x - 2) = 2\).

  1. A\(x = 5\)
  2. B\(x = -7\) or \(x = 3\)
  3. C\(x = 3\)
  4. D\(x = 3\) or \(x = 7\)
  5. E\(x = -3\)
  6. F\(x = 1\)
Answer: C. \((x+6)(x-2) = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\). \(x = -7\) makes both logs undefined, so \(x = 3\) only.
Full method — how and why
Combine the logs, convert to an equation, then check the domain - that is what this question is really testing.
  1. Combine: \(\log_3\left[(x+6)(x-2)\right] = 2\).
    Sum of logs is the log of the product.
  2. Convert to exponential form: \((x+6)(x-2) = 3^2 = 9\).
    \(\log_a N = c\) means \(N = a^c\).
  3. Expand and solve: \(x^2 + 4x - 12 = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\), so \(x = -7\) or \(x = 3\).
    Ordinary quadratic work.
  4. Check both in the original equation: \(x = 3\) gives \(\log_3 9 + \log_3 1\) ✓; \(x = -7\) gives \(\log_3(-1)\), which is undefined. So \(x = 3\) only.
    Logarithms of negative numbers (or zero) do not exist, so roots that break the original expression must be rejected.
Where it goes wrong
  • Giving both roots - the option '\(x = -7\) or \(x = 3\)' exists precisely to catch this.
  • Checking only one of the two logarithms.
Takeaway Always substitute your solutions back: combining logs can create solutions the original equation never had.
6. LOG-07Exponentials and logarithms · spec MM5.2 · Paper 2 style · ★☆☆

Which of the following statements is/are true for all positive real \(x\) and \(y\)?

  1. I\(\log_{10}(x + y) = \log_{10} x + \log_{10} y\)
  2. II\(\log_{10}(xy)^2 = 2\log_{10} x + 2\log_{10} y\)
  3. III\(\log_{10} x - \log_{10} y = \log_{10} x \div \log_{10} y\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: C. II is correct: \(\log(xy)^2 = 2\log(xy) = 2\log x + 2\log y\). I and III confuse the laws: \(\log\) of a sum has no simple form, and a difference of logs is \(\log\) of a quotient, not a quotient of logs.
Full method — how and why
Three log 'laws', two of which are fabrications. Test each against the genuine rules.
  1. I: \(\log(x+y)\) has no expansion. Test \(x = y = 1\): \(\log 2 \approx 0.30\) but \(\log 1 + \log 1 = 0\). False.
    The product law is \(\log(xy) = \log x + \log y\); the sum inside is a different animal entirely.
  2. II: \(\log(xy)^2 = 2\log(xy) = 2(\log x + \log y) = 2\log x + 2\log y\). True.
    Power law first, then product law - both applied correctly.
  3. III: \(\log x - \log y = \log\frac xy\), which is not the same as \(\frac{\log x}{\log y}\). Test \(x = 100\), \(y = 10\): left \(= 1\), right \(= \frac21 = 2\). False.
    Difference of logs is the log of a quotient, not the quotient of logs.
  4. II only.
    Match to the options.
Where it goes wrong
  • Accepting I by analogy with the product law.
  • Accepting III because 'subtraction corresponds to division' - it does, but inside the log.
Takeaway The laws act on what is inside the log: \(\log(xy)\), \(\log\frac xy\), \(\log x^k\). Nothing simplifies \(\log(x+y)\).
7. LOG-09Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Find the value of \(\dfrac{\log_5 27}{\log_5 3}\).

  1. A\(9\)
  2. B\(\dfrac{27}{3}\)
  3. C\(\log_5 9\)
  4. D\(3\)
  5. E\(\log_5 24\)
  6. F\(24\)
Answer: D. \(\log_5 27 = \log_5 3^3 = 3\log_5 3\), so the quotient is \(3\).
Full method — how and why
A quotient of two logs with the same base. Rewriting the numerator's argument as a power makes it cancel.
  1. Write \(27 = 3^3\), so \(\log_5 27 = \log_5 3^3 = 3\log_5 3\).
    The power law applies to the argument, whatever the base of the log.
  2. Form the quotient: \(\dfrac{3\log_5 3}{\log_5 3} = 3\).
    The common factor \(\log_5 3\) cancels (it is non-zero).
Where it goes wrong
  • Subtracting to get \(\log_5 24\) - a quotient of logs is not the log of a quotient.
  • Dividing the arguments to get \(\frac{27}{3} = 9\) in log form.
Takeaway \(\frac{\log_c a}{\log_c b} = \log_b a\): quotients of logs change the base, they do not divide the arguments.
8. LOG-10Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

How many real solutions does the equation \(4^{x} + 2^{x+1} - 8 = 0\) have?

  1. A\(1\)
  2. B\(2\)
  3. C\(0\)
  4. D\(3\)
  5. Einfinitely many
Answer: A. \(y = 2^x\): \(y^2 + 2y - 8 = (y + 4)(y - 2) = 0\). \(y = -4\) is impossible, \(y = 2\) gives \(x = 1\). One solution.
Full method — how and why
Quadratic in \(2^x\) again, but this time one root must be rejected - which changes the count.
  1. Rewrite: \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x+1} = 2\cdot 2^x\).
    Get everything in terms of \(2^x\) before substituting.
  2. Let \(y = 2^x\): \(y^2 + 2y - 8 = 0\), which factorises as \((y+4)(y-2) = 0\).
    Standard quadratic.
  3. Reject \(y = -4\): \(2^x\) is positive for every real \(x\), so it can never equal a negative number.
    This is the crucial filter - exponentials are always strictly positive.
  4. From \(y = 2\): \(2^x = 2\), so \(x = 1\). Exactly one real solution.
    The question asks how many, so the count is the answer.
Where it goes wrong
  • Answering \(2\) because the quadratic has two roots.
  • Rejecting \(y = 2\) as well, and answering \(0\).
Takeaway After substituting \(y = a^x\), discard every root with \(y \leq 0\) - then count what remains.
9. LOG-11Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_2 x + \log_4 x = 6\). (You may use \(\log_4 x = \dfrac{1}{2}\log_2 x\).)

  1. A\(x = 8\)
  2. B\(x = 2^{6}\)
  3. C\(x = 16\)
  4. D\(x = 64\)
  5. E\(x = 4\)
  6. F\(x = 32\)
Answer: C. \(\frac32\log_2 x = 6 \Rightarrow \log_2 x = 4 \Rightarrow x = 16\).
Full method — how and why
Two different bases. The hint converts one to the other, after which it is a linear equation in \(\log_2 x\).
  1. Use the given relation \(\log_4 x = \frac12\log_2 x\).
    It holds because \(4 = 2^2\): it takes twice as many factors of \(2\) as of \(4\) to build \(x\).
  2. Substitute: \(\log_2 x + \frac12\log_2 x = 6\), i.e. \(\frac32\log_2 x = 6\).
    Both terms now involve the same quantity, so treat \(\log_2 x\) as a single unknown.
  3. Solve for the log: \(\log_2 x = 4\).
    Multiply by \(\frac23\).
  4. Convert: \(x = 2^4 = 16\).
    \(\log_2 x = 4\) means \(x = 2^4\).
Where it goes wrong
  • Solving \(\log_2 x = 6\) and answering \(64\) - which is offered (twice, as \(64\) and \(2^6\)).
  • Adding the bases or the arguments instead of using the relation.
Takeaway Get every logarithm to the same base first; then it is ordinary algebra in that one quantity.
10. LOG-12Exponentials and logarithms · spec MM5.1 · Paper 1 style · ★☆☆

The curve \(y = a^{x}\) passes through \((2, 12)\) and \((4, 108)\). Find \(a\).

  1. A\(4\)
  2. B\(3\)
  3. C\(\sqrt{12}\)
  4. D\(9\)
  5. E\(2\)
  6. F\(6\)
Answer: B. \(\frac{a^4}{a^2} = a^2 = \frac{108}{12} = 9\), so \(a = 3\) (as \(a > 0\)).
Full method — how and why
Two points on an exponential curve. Dividing eliminates any constant and isolates a power of \(a\).
  1. Write both facts: \(a^2 = 12\) and \(a^4 = 108\).
    Substitute each point into \(y = a^x\).
  2. Divide: \(\dfrac{a^4}{a^2} = a^2 = \dfrac{108}{12} = 9\).
    Dividing subtracts indices, giving a single power of \(a\) - the standard move with two points.
  3. Solve: \(a = 3\) (bases of exponentials are positive).
    Reject \(a = -3\): \(y = a^x\) requires \(a > 0\).
Where it goes wrong
  • Answering \(9\) - that is \(a^2\), not \(a\).
  • Subtracting the \(y\)-values instead of dividing.
Takeaway Divide two points on an exponential curve; the index difference gives a clean power of the base.
11. LOG-13Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Given that \(\log_{10} 2 = p\) and \(\log_{10} 3 = q\), express \(\log_{10} 0.75\) in terms of \(p\) and \(q\).

  1. A\(3q - 4p\)
  2. B\(\dfrac{q}{2p}\)
  3. C\(2p - q\)
  4. D\(q + 2p\)
  5. E\(q - p\)
  6. F\(q - 2p\)
Answer: F. \(0.75 = \frac34\), so \(\log 0.75 = \log 3 - \log 4 = q - 2p\).
Full method — how and why
Express the number in terms of the given primes, then apply the log laws.
  1. Write \(0.75 = \dfrac34\).
    Decimals must become fractions of small numbers before the laws help.
  2. Split: \(\log_{10}\frac34 = \log_{10}3 - \log_{10}4\).
    Quotient law.
  3. Write \(4 = 2^2\): \(\log_{10}4 = 2\log_{10}2 = 2p\).
    Only \(\log 2\) and \(\log 3\) are available, so every number must be built from \(2\)s and \(3\)s.
  4. Combine: \(q - 2p\).
    Substituting \(\log_{10}3 = q\).
Where it goes wrong
  • Using \(\log 4 = p^2\) or \(= 2 + p\).
  • Getting the subtraction backwards and answering \(2p - q\).
Takeaway Factorise the argument into the primes you have been given, then apply product/quotient/power laws.
12. LOG-15Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_2 (x^2 - 4) - \log_2 (x - 2) = 3\).

  1. A\(x = 2\)
  2. B\(x = 6\)
  3. C\(x = -10\) or \(x = 6\)
  4. D\(x = 8\)
  5. E\(x = 10\)
  6. F\(x = 4\)
Answer: B. For \(x > 2\): \(\log_2\frac{(x-2)(x+2)}{x-2} = \log_2(x + 2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).
Full method — how and why
Combine the logs; the algebraic fraction simplifies dramatically because the numerator factorises.
  1. Combine: \(\log_2\dfrac{x^2-4}{x-2} = 3\).
    Difference of logs is the log of the quotient.
  2. Factorise and cancel: \(\dfrac{(x-2)(x+2)}{x-2} = x + 2\) (valid since \(x \neq 2\)).
    The domain requires \(x > 2\) for \(\log_2(x-2)\) to exist, so cancelling is legitimate.
  3. Convert: \(\log_2(x+2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).
    \(2^3 = 8\).
  4. Check the domain: \(x = 6\) gives \(x^2 - 4 = 32 > 0\) and \(x - 2 = 4 > 0\) ✓.
    Always verify against the original.
Where it goes wrong
  • Forgetting the domain and including a spurious negative solution.
  • Cancelling incorrectly to get \(x - 2\) or \(x^2 - 2\).
Takeaway Factorise before cancelling inside a log, and confirm every solution keeps all arguments positive.
13. LOG-16Exponentials and logarithms · spec MM5.1 · Paper 2 style · ★☆☆

For which of the following values of \(x\) is \(3^{x} < 2^{x+1}\)?

  1. I\(x = 0\)
  2. II\(x = 1\)
  3. III\(x = 2\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: E. \(x = 0\): \(1 < 2\) true. \(x = 1\): \(3 < 4\) true. \(x = 2\): \(9 < 8\) false. So I and II.
Full method — how and why
Three numerical tests. Nothing conceptual - just careful arithmetic with small powers.
  1. I: \(x = 0\) gives \(3^0 = 1\) and \(2^{1} = 2\). Is \(1 < 2\)? Yes. True.
    \(a^0 = 1\) for any \(a\).
  2. II: \(x = 1\) gives \(3^1 = 3\) and \(2^2 = 4\). Is \(3 < 4\)? Yes. True.
    Careful with the index \(x + 1 = 2\).
  3. III: \(x = 2\) gives \(3^2 = 9\) and \(2^3 = 8\). Is \(9 < 8\)? No. False.
    \(3^x\) has overtaken \(2^{x+1}\) by here - exponential growth with the larger base always wins eventually.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Computing \(2^{x+1}\) as \(2^x + 1\).
  • Assuming the inequality holds for all \(x\) because it holds for the first two.
Takeaway Substitute each value carefully; a larger base always overtakes eventually, whatever the head start.
14. LOG-18Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

If \(\log_a b = 2\) and \(\log_b c = 3\), find \(\log_a c\).

  1. A\(8\)
  2. B\(6\)
  3. C\(\dfrac{2}{3}\)
  4. D\(9\)
  5. E\(5\)
  6. F\(\dfrac{3}{2}\)
Answer: B. \(b = a^2\) and \(c = b^3 = a^6\), so \(\log_a c = 6\).
Full method — how and why
A chain of logarithms. Convert each statement to exponential form and substitute.
  1. \(\log_a b = 2\) means \(b = a^2\).
    \(\log_a b = c \Leftrightarrow a^c = b\).
  2. \(\log_b c = 3\) means \(c = b^3\).
    Same conversion with the next pair.
  3. Substitute: \(c = (a^2)^3 = a^6\).
    Powers of powers multiply.
  4. Therefore \(\log_a c = 6\).
    Read the exponent straight off. (In general \(\log_a c = \log_a b \times \log_b c = 2 \times 3\).)
Where it goes wrong
  • Adding the values to get \(5\).
  • Reading \((a^2)^3\) as \(a^5\).
Takeaway Logarithms chain by multiplication: \(\log_a c = \log_a b \times \log_b c\).
15. LOG-06Exponentials and logarithms · spec MM5.1 · Paper 1 style · ★★☆

The graph of \(y = 2^{x}\) is translated \(3\) units in the positive \(x\)-direction. The resulting graph has equation

  1. A\(y = \dfrac{1}{8}\cdot 2^{x}\)
  2. B\(y = 2^{x/3}\)
  3. C\(y = 8\cdot 2^{x}\)
  4. D\(y = 2^{x} - 3\)
  5. E\(y = 2^{3x}\)
  6. F\(y = 2^{x} + 3\)
Answer: A. \(y = 2^{x - 3} = 2^{-3}\cdot 2^x = \frac18 \cdot 2^x\). (A translation of an exponential is also a stretch.)
Full method — how and why
A translation of an exponential graph is also a stretch. The algebra shows why.
  1. Translating \(3\) right replaces \(x\) by \(x - 3\): \(y = 2^{x-3}\).
    A shift to the right by \(a\) substitutes \(x - a\).
  2. Split the index: \(2^{x-3} = 2^x \times 2^{-3} = \dfrac{1}{8}\cdot 2^x\).
    \(a^{m-n} = \frac{a^m}{a^n}\) - the translation has become multiplication by a constant.
  3. So the new graph is the old one stretched vertically by factor \(\frac18\).
    This is special to exponentials: horizontal translation and vertical stretch are the same thing.
Where it goes wrong
  • Writing \(2^{x-3} = 2^x - 3\) and choosing that option.
  • Using \(2^{+3} = 8\) as the factor, which corresponds to a shift to the left.
Takeaway \(a^{x-k} = a^{-k}a^x\): for exponentials, translating horizontally is the same as stretching vertically.
16. LOG-08Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

Given that \(5^{x} = 2\) and \(2^{y} = 125\), find the value of \(xy\).

  1. A\(1\)
  2. B\(3\)
  3. C\(5\)
  4. D\(2\)
  5. E\(6\)
  6. F\(\log_5 2 \cdot \log_2 125\)
Answer: B. \(5^{xy} = (5^x)^y = 2^y = 125 = 5^3\), so \(xy = 3\).
Full method — how and why
Two exponential facts that chain together. Raising the first to the power \(y\) links them.
  1. Raise the first equation to the power \(y\): \((5^x)^y = 2^y\), i.e. \(5^{xy} = 2^y\).
    \((a^x)^y = a^{xy}\) - this is what connects the two statements.
  2. Substitute the second fact: \(2^y = 125\).
    Given directly.
  3. Write \(125\) as a power of \(5\): \(5^{xy} = 5^3\).
    Both sides are now powers of \(5\).
  4. Equate the indices: \(xy = 3\).
    Valid because \(5^u\) is one-to-one.
Where it goes wrong
  • Trying to compute \(x\) and \(y\) separately as logarithms and getting stuck without a calculator.
  • Multiplying the two equations to get \(10^{x+y} = 250\), which leads nowhere.
Takeaway To find a product of exponents, raise one equation to the other's exponent and chain the results.
17. LOG-14Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

The number of integers \(n\) satisfying \(2^{n} < 1000 < 2^{n+3}\) is

  1. A\(3\)
  2. B\(4\)
  3. C\(0\)
  4. D\(1\)
  5. E\(10\)
  6. F\(2\)
Answer: A. \(2^9 = 512 < 1000 < 1024 = 2^{10}\). Need \(n \leq 9\) and \(n + 3 \geq 10\), so \(n = 7, 8, 9\): three integers.
Full method — how and why
A double inequality in powers of \(2\). Locate \(1000\) between consecutive powers, then translate into conditions on \(n\).
  1. Find where \(1000\) sits: \(2^9 = 512\) and \(2^{10} = 1024\), so \(2^9 < 1000 < 2^{10}\).
    Powers of \(2\) up to \(1024\) are worth knowing by heart.
  2. The left condition \(2^n < 1000\) requires \(n \leq 9\).
    \(2^{10} = 1024 > 1000\), so \(n = 10\) fails.
  3. The right condition \(1000 < 2^{n+3}\) requires \(n + 3 \geq 10\), i.e. \(n \geq 7\).
    \(2^{9} = 512 < 1000\), so \(n + 3 = 9\) (i.e. \(n = 6\)) fails.
  4. Combine: \(7 \leq n \leq 9\), giving \(n = 7, 8, 9\) - three integers.
    Count the integers in the range, inclusive.
Where it goes wrong
  • Counting \(9 - 7 = 2\) instead of \(3\).
  • Using strict inequalities on \(n\) where the powers allow equality.
Takeaway Convert conditions on \(2^n\) into conditions on \(n\) by comparing with the nearest powers, then count inclusively.
18. LOG-17Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

The equation \(9^{x} - k\cdot 3^{x} + 9 = 0\) has exactly one real solution. Find the value of \(k\).

  1. A\(3\)
  2. B\(-6\)
  3. C\(\pm 6\)
  4. D\(9\)
  5. E\(6\)
  6. F\(0\)
Answer: E. With \(y = 3^x > 0\): \(y^2 - ky + 9 = 0\). Exactly one solution for \(x\) means exactly one positive root: a repeated root (\(k^2 = 36\), product \(9 > 0\) so the roots have the sign of \(k\)): \(k = 6\) (repeated root \(y = 3\), \(x = 1\)). \(k = -6\) gives \(y = -3\), no solutions. (Two distinct positive roots would give two solutions.)
Full method — how and why
A quadratic in \(3^x\) with a parameter. 'Exactly one real solution' is a condition on the positive roots, not simply on the discriminant.
  1. Substitute \(y = 3^x\), noting \(y > 0\): \(y^2 - ky + 9 = 0\).
    \(9^x = (3^x)^2\). The restriction \(y > 0\) is the heart of the question.
  2. The product of the roots is \(9 > 0\), so the roots have the same sign; their sum is \(k\), so both are positive exactly when \(k > 0\).
    Sum and product of roots decide the signs without solving.
  3. For exactly one value of \(x\), we need exactly one positive root, which (given both roots share a sign) means a repeated root: discriminant \(k^2 - 36 = 0\), so \(k = \pm6\).
    Two distinct positive roots would give two solutions for \(x\); two negative roots would give none.
  4. Check the signs: \(k = 6\) gives the repeated root \(y = 3 > 0\), so \(x = 1\) - exactly one solution. \(k = -6\) gives \(y = -3 < 0\), so no solutions at all. Hence \(k = 6\).
    The final sign check is what rules out \(-6\), which is offered.
Where it goes wrong
  • Answering \(\pm6\) from the discriminant alone.
  • Forgetting that \(y\) must be positive and counting \(y = -3\) as a solution.
Takeaway With \(y = a^x\), translate conditions on \(x\) into conditions on the positive roots in \(y\), using sum and product to check signs.
19. LOG-19Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★★☆

A quantity halves every \(5\) hours. Which expression gives the number of hours for it to fall to \(\dfrac{1}{10}\) of its initial value?

  1. A\(5\log_2 5\)
  2. B\(\dfrac{\log_{10} 2}{5}\)
  3. C\(5\log_{10} 2\)
  4. D\(\dfrac{5}{\log_{10} 2}\)
  5. E\(25\)
  6. F\(50\log_{10} 2\)
Answer: D. \(\left(\frac12\right)^{t/5} = \frac{1}{10}\). Taking logs to base \(10\): \(-\frac t5 \log_{10} 2 = -1\), so \(t = \dfrac{5}{\log_{10} 2}\) (about \(16.6\) hours).
Full method — how and why
Exponential decay. Build the model, then take logs - and be careful which expression the answer takes.
  1. Model: after \(t\) hours the quantity is multiplied by \(\left(\frac12\right)^{t/5}\), because \(t/5\) half-lives have passed.
    Dividing \(t\) by the half-life counts how many halvings have occurred.
  2. Set it equal to \(\frac{1}{10}\): \(\left(\frac12\right)^{t/5} = \dfrac{1}{10}\).
    'Falls to one tenth' is a statement about the multiplier.
  3. Take \(\log_{10}\): \(\dfrac t5\log_{10}\dfrac12 = \log_{10}\dfrac{1}{10} = -1\), i.e. \(-\dfrac t5\log_{10}2 = -1\).
    \(\log\frac12 = -\log 2\) and \(\log\frac1{10} = -1\); the two minus signs will cancel.
  4. Solve: \(t = \dfrac{5}{\log_{10}2}\) (about \(16.6\) hours).
    Multiply both sides by \(-\frac{5}{\log_{10}2}\).
Where it goes wrong
  • Answering \(5\log_{10}2\) - the reciprocal, which would be under a second.
  • Sanity check: the quantity halves in \(5\) h and \(\frac1{10} < \frac18\), so \(t\) must exceed \(15\) h; \(5\log_{10}2 \approx 1.5\) is obviously wrong.
Takeaway Set up the multiplier, take logs, and sanity-check the size of the answer against a couple of half-lives.
20. X2-06Exponentials and logarithms · spec MM5.2 · Paper 2 style · ★★☆

Which of the following is a counterexample to the claim 'for all positive \(x\) and \(y\), \(\log_{10}(x + y) \leq \log_{10} x + \log_{10} y\)'?

  1. A\(x = 10\), \(y = 100\)
  2. B\(x = y = 2\)
  3. C\(x = 5\), \(y = 5\)
  4. D\(x = 4\), \(y = 4\)
  5. E\(x = y = 1\)
Answer: E. \(x = y = 1\): \(\log 2 > 0 = \log 1 + \log 1\), so the inequality fails. For the others \(x + y \leq xy\) and the claim holds (\(\log\) is increasing).
Full method — how and why
To break an inequality claim, look for the case where the two sides are closest - here small values, where logs go to zero or negative.
  1. Test \(x = y = 1\): left side \(\log_{10}2 \approx 0.30\); right side \(\log_{10}1 + \log_{10}1 = 0\).
    \(\log 1 = 0\) for any base, so the right side collapses.
  2. \(0.30 > 0\), so the claimed inequality \(\leq\) fails. Counterexample.
    Hypothesis satisfied (\(x,y\) positive), conclusion false.
  3. Why the others work: \(\log x + \log y = \log(xy)\), and \(\log\) is increasing, so the claim is just \(x+y \leq xy\). For \(x=y=2\): \(4 \leq 4\) ✓; \(x=y=4\): \(8 \leq 16\) ✓; \(x=y=5\): \(10\leq25\) ✓; \(x=10,y=100\): \(110 \leq 1000\) ✓.
    Converting the log inequality into an algebraic one makes all five cases quick to check.
Where it goes wrong
  • Trying large values, where \(xy\) dwarfs \(x+y\) and the claim comfortably holds.
  • Forgetting \(\log_{10}1 = 0\) and miscomputing the right side.
  • Thinking \(\log(x+y) = \log x + \log y\) - that is the error the whole question is built on.
Takeaway \(\log x + \log y = \log(xy)\), never \(\log(x+y)\). Test inequality claims near the boundary, where the sides are closest.
21. X2-22Exponentials and logarithms · spec MM5.3 · Paper 2 style · ★★☆

Consider the statement: for all real \(x\), \(2^{x} > x^2\). Which of the following is a counterexample?

  1. A\(x = 5\)
  2. B\(x = 3\)
  3. C\(x = 0\)
  4. D\(x = 10\)
  5. E\(x = 1\)
Answer: B. \(2^3 = 8 < 9 = 3^2\), so \(x = 3\) breaks the claim. The other options satisfy it: \(2 > 1\), \(32 > 25\), \(1 > 0\), \(1024 > 100\).
Full method — how and why
Exponential versus quadratic growth: the exponential eventually wins, but there is a window where it loses. Test values inside that window.
  1. Check the candidates: \(x=0\) gives \(1>0\) ✓; \(x=1\) gives \(2>1\) ✓; \(x=5\) gives \(32>25\) ✓; \(x=10\) gives \(1024>100\) ✓.
    Evaluate both sides for each - no cleverness needed.
  2. \(x=3\): \(2^3 = 8\) and \(3^2 = 9\). Since \(8 < 9\), the claim fails. Counterexample.
    Hypothesis (\(x\) real) satisfied, conclusion false.
  3. Why here? \(2^x\) and \(x^2\) cross at \(x=2\) (both \(4\)) and again near \(x=4\) (both \(16\)), so the quadratic is ahead between them.
    Knowing where the curves cross tells you exactly where to hunt.
Where it goes wrong
  • Testing only large \(x\), where the exponential dominates and the claim holds.
  • Overlooking the window: the two sides are equal at \(x=2\) and \(x=4\), so the quadratic leads strictly between them - that is where a counterexample must lie.
  • Assuming exponentials always beat polynomials - true only eventually.
Takeaway \(2^x\) and \(x^2\) meet at \(x=2\) and \(x=4\); between them the quadratic is larger. 'Eventually bigger' is not 'always bigger'.